Samantha from
Hamlin sent us her work on this problem. She's assumed that each
letter has to represent a different number.
There are two solutions to this problem.
I knew that F had to equal 0 because R minus something could only
equal R if you were subtracting 0.
I then looked at the letter combination in the 2nd and 4th
columns. They both contained the letters ALE, but arranged in
different orders. I knew that A had to be bigger than L and E
because in the second column, A can't borrow from R. That meant
that in the last column, E had to borrow from R. Therefore, since
R minus E equals E, R had to be one more than 2E, because you are
borrowing one from it. That meant that R could be 9, 7, 5, or 3.
I decided to try using seven first, which would mean that E= 3.
So far, I had
| 7 | A | 7 | 3 |
| - | 0 | L | 3 | A |
| - | - | - | - | - |
| 7 | 3 | 3 | L
|
I then knew that A and L had to be three apart. The only three
possible combinations were 9 and 6, 8 and 5, and 4 and 1
because all of the other combinations used numbers that we had
already used or were not three apart. I also knew that A and L
had to equal 13. 8+5=13.
I then filled in the other letters. I had
| 7 | 8 | 7 | 3 |
| - | 0 | 5 | 3 | 8 |
| - | - | - | - | - |
| 7 | 3 | 3 | 5
|
One answer is R=7, A=8, E=3, L=5, and F=0.
I then needed to make sure that there was only one solution. I
tried substituting 9 for R. That would mean that E was 4. I
needed two numbers for A and L that were 4 apart and added up
to 14. The only combinations were 9 and 5, 8 and 4, 7 and 3, 6
and 2, and 5 and 1. The first two pairs were eliminated because
they used numbers that had been previously used (R=9 and E=4).
None of the remaining number pairs added up to 14.
I then tried substituting 5 for R. That would mean that E was
2. I needed two numbers for A and L that were 2 apart and added
up to 12. The only combinations were 9 and 7, 8 and 6, 7 and 5,
6 and 4, 5 and 3, 4 and 2 and 3 and 1. 7 and 5, 5 and 3, and 4
and 2 were eliminated because they used numbers that had been
previously used (R=5 and E=2). None of the remaining number
pairs added up to 12.
Finally, I tried substituting 3 for R. That would mean that E
was 1. I needed two numbers for A and L that were 1 apart and
added up to 11. The only combinations were, 9 and 8, 8 and 7, 7
and 6, 6 and 5, 5 and 4, 4 and 3, 3 and 2, and 2 and 1. The
last two solutions were eliminated because they used numbers
that had been previously used (R=2 and E=1). However, 6 and 5
added up to 11. I then substituted the remaining numbers. I
came up with
| 3 | 6 | 3 | 1 |
| - | 0 | 5 | 1 | 6 |
| - | - | - | - | - |
| 3 | 1 | 1 | 5
|
The second solution is R=3, A=6, E=1, L=5, and F=0.
Good work, Samantha!